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I hope someone can help me.

I have this green function f (a sigmoid), and I would like to have the equation of the blue function that is the "mirror" of f by the red function.

I have no idea how to achieve that.

Can someone help me please ?

enter image description here

Filimindji
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    Apart from the shift by 0.5 in y-direction, you are looking for the inverse function of $f$. A simple calculation gives $g(x) = \frac 1a \cdot \ln \left( \frac{1+2x}{1-2x} \right) + \frac 12$: https://www.desmos.com/calculator/qaoklvaby7. – Martin R Nov 23 '19 at 19:53
  • Hey. I didn't know that $ \log(x) \cdot \ln(10) = \ln(x) $ . It's great, thank you. – Filimindji Nov 24 '19 at 14:09
  • Inverse of sigmoid is logit and you can see here the step by step https://math.stackexchange.com/questions/3480014/logit-to-sigmoid-explanation – Mr-Programs Dec 18 '19 at 13:27

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It looks similar to $\arccos(-x)$ but for the sigmoid funtion its inverse is denoted as logit fuction maybe with some scaling and translations easy to apply.

user
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  • Hey, it's pretty close, thank you : http://prntscr.com/q11i1f – Filimindji Nov 23 '19 at 16:22
  • @Filimindji Ah you have used $\arccos (-x)$. And what about logit function? – user Nov 23 '19 at 16:26
  • I'm looking how to use the logit, I cant yet figure how to use the alpha parameter from the sigmoid : log( xalpha / (1 - xalpha)) do not give the result I expected. – Filimindji Nov 23 '19 at 16:32
  • Not sure how I came to this approximation, but this work well : http://prntscr.com/q11rhl (sorry about that, I'm pretty sure it's horible to put random value in the equation to just make it look like what I want) – Filimindji Nov 23 '19 at 16:46
  • Yes it looks fine! – user Nov 23 '19 at 16:47
  • Looks like it's better to use ln(10) instead of the approximation 2.3 I used. So far : http://prntscr.com/q1239d. I'm not really happy with the m - x at the end of the equation, but still, it fit my needs – Filimindji Nov 23 '19 at 17:15